POJ - 3264 Balanced Lineup

Posted 2020-09-30 -凡-尘

题目链接：http://poj.org/problem?id=3264

题目大意：

一个农夫有N头牛，每头牛的高度不同，我们需要找出最高的牛和最低的牛的高度差。

解题思路：经典 RMQ

#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
#define rep(i,a,b) for(int i=a;i<=b;i++)
int n,q,l,r;
int a[50005];
int maxx[50005][25],minn[50005][25];

void RMQ()
{
    rep(j,1,20)
        rep(i,1,n)
            if(i + (1 << j)-1<=n)
            {
                maxx[i][j] = max(maxx[i][j-1], maxx[i + (1 << (j - 1))][j - 1]);
                minn[i][j] = min(minn[i][j-1], minn[i + (1 << (j - 1))][j - 1]);
            }
}
int main()
{
    while(~scanf("%d%d",&n,&q))
    {
        rep(i,1,n)
        {
            scanf("%d",&a[i]);
            maxx[i][0]=minn[i][0]=a[i];
        }
        RMQ();

        rep(i,1,q)
        {
            scanf("%d%d",&l,&r);
            int k=(int)(log(r-l+1.0)/log(2.0));
            int maxnum = max(maxx[l][k], maxx[r - (1 << k) +1][k]);
            int minnum = min(minn[l][k], minn[r - (1 << k) +1][k]);
            //printf("%d %d\n",maxnum,minnum);
            printf("%d\n", maxnum - minnum);
        }
    }
    return 0;
}